Given a sorted array nums, remove the duplicates in-place such that each element appear only once and return the new length.

Do not allocate extra space for another array, you must do this by modifying the input array in-place with O(1) extra memory.

Example 1:

Given nums = [1,1,2],

Your function should return length = 2, with the first two elements of nums being 1 and 2 respectively.

It doesn’t matter what you leave beyond the returned length.
Example 2:

Given nums = [0,0,1,1,1,2,2,3,3,4],

Your function should return length = 5, with the first five elements of nums being modified to 0, 1, 2, 3, and 4 respectively.

It doesn’t matter what values are set beyond the returned length.
Clarification:

Confused why the returned value is an integer but your answer is an array?

Note that the input array is passed in by reference, which means modification to the input array will be known to the caller as well.

Internally you can think of this:

// nums is passed in by reference. (i.e., without making a copy)
int len = removeDuplicates(nums);

// any modification to nums in your function would be known by the caller.
// using the length returned by your function, it prints the first len elements.
for (int i = 0; i < len; i++) {
    print(nums[i]);
}
 

題目解析：這道題目要求我們去除有序陣列中重複的數字，使陣列中的每一個數字元素只出現一次，返回的是改變後陣列的長度。

思路：1、我們通過設定兩個指標，一個快指標，一個慢指標，當出現數據相等的情況，快指標向後移動，慢指標不變，不等時，快慢指標均向後移動。

class Solution {
public:
    int removeDuplicates(vector<int>& nums) {
        if(nums.empty()) return 0;
        int pre=0,cur=0,n=nums.size();
        //快慢指標，快指標為cur,慢指標為pre
        while(cur<n){
            if(nums[pre]==nums[cur])++cur;
            //如果存在前後相等，則只讓當快指標向後移動，不移動慢指標
            else nums[++pre]=nums[cur++];
            //否則均向後移動
            
        }
        
        return pre+1;
    }
};

2、利用vector中的函式erase，來刪除出現的重複數字。

class Solution
{
public:
	int removeDuplicates(vector<int> & nums)
	{	
		int i;
		for(i=1;i!=num.size();i++)
		{
			if(nums[i]==nums[i-1])
				{
					nums.erase(nums.begin()+i);
					i--;
				}
		}
		return nums.size();
	}
};